Hi, welcome to module 29 of two dimensional dynamics. Today's learning outcome is to solve an engineering problem for the kinetics of a rigid body translating in two dimensional planar motion. So, this is the theory we develop last time. We came up with the equations. And we're going to now apply this theory to this problem. And I'm going to change the nomenclature here slightly. So that we don't get confused. I'm going to call the force F. And then to be consistent with my theory I'm going to call this point of contact P. And so, I want to find the angle for this bar for which the bar will translate to the right with a constant in the angular ex, or excuse me, a constant acceleration. And then I also want to find what force is required to maintain that translation. So let's look at a demo. I always like to Look at the real world or physical significance of these problems. So here is my, my, my bar. I've got a, a surface with a coefficient of friction u. I've got a force f down here and I want to push to the, to the to the right where it stays at a constant angle and a constant acceleration. And I'm not quite doing it but I'm getting close. And so we want it to go along with the constant acceleration or translation. And so my first question to you is, how do we approach this problem? And how are you going to start? And what you should do, this is often the most difficult part of the problem to, to set it up. And what you want to do is here, I like to use these graphical tools, like I said in statics, you always want to draw a free body diagram. Let's draw a free body diagram but now we have kinetic or motion vectors in the right hand side in the equation, so i'm also going to use a kinetic diagram. And so, let's go ahead and have you begin by doing the free body diagram and come on back and see how you did. And so here's the free body diagram. You can see if you you got that correct. Let's now proceed with the kinetic diagram. So, here's my body of interest, the bar. Here is the mass center, C, and I'm going to put my kinetic vectors or motion vectors At the mass center ma is the vector. I'll break it into two orthogonal components. I'll use the X and Y directions So I'm going to have an max and an may. Can we simplify that kinetic diagram at all? And what you should say is yes. The [UNKNOWN] is translating maY should be equal to zero. And so let's go back and look at my demonstration. And again, if this is translating, then the acceleration of the mass center is only In the horizontal direction. There's no. >> [SOUND] >> Acceleration in the vertical direction. And so that's a simplification. So we'll just go ahead and call max, we'll just call it ma. Since there's only one component of acceleration. All right, let's pick up from there, and here's my free body versus and kinetic diagram, free body diagram and kinetic diagram, what do I do next? And so, what you should say is I'm going to go ahead and apply my equations, I can either sum moments about P or C, P being any arbitrary point. I'll actually pick point P. That's sort of a smart point, if you will, because it gets rid of three of my unknowns. All three of these forces have a line of action through point P, so they're not going to cause a moment about point P. So, let's write the equation, some of the moments about P on the free body diagram. Equals sum of the moments. About P on the kinetic diagram. I'll choose counterclockwise positive for sum for assembling my equation. And so if I sum moments about P on my free body diagram, the only force that's going to cause a moment is the weight force, mg. It'll tend to cause a clockwise rotation, which is opposite my positive sin convention. So it's going to be minus the force mg times its moment arm. Which is going to be l over two cosin theta. Equals, that's the only force causing ten, with a tendency to cause a, a rotation or moment about P on the free body diagram. Over here on the kinetic diagram, I've got my MA vector or affective force. It's going to cause a, going to cause a clockwise rotation. And so that's again negative in this accordance with my sine convention. It's moment arm will be L/2sinTheta Then I can simplify this equation by canceling the m's, cancelling l over two. And I get g over a equals sin theta over cosin theta. But we all know that sin theta over cosin theta is tangent theta. And so Are results is that theta equals the tangent inverse of G over A, and I've now solved one portion of the problem. I have now found the angle for which the bar translates to the right with a constant acceleration. Whatever that acceleration is, if I divide that into my acceleration due to gravity, take the inverse tangent, that'll tell me what my angle is. And so let's continue on. There's the result so far. Next I want to find the force F required to maintain that translation. How would I do that? And what you should say is well, let's go ahead and Do another equation for forces in the X direction, since we want to find that force F. And so I've got sum of the forces in the F direction on my free body diagram, equals sum of the forces in the, X, X direction on my kinetic diagram. I'll choose to the right positive. And so I get F - uN = ma, or F=ma+uN. Alright, let's use that result. So, so far we've got theta, and we've got the force, but we need to put force we have to find out what the normal force is, and so to do that we'll go ahead and sum forces in the Y direction for both my free body diagram and kinetic diagram, and some of the forces in the Y Of the free body diagram equals sum of the forces in the Y. In the kinetic diagram, I'll choose up positive, and so I get N-mg. N is up, positive. MG is down, negative equals no kinetic vectors in the Y direction, 0. So N, Equals mg. And then go ahead and substitute that in up here. And I arrive at my force equals ma plus u times mg. And I can factor out the m. So I get F equals m times ug plus a. And so the force required will be whatever the mass in the bar is times the quantity, the coefficient of friction times the acceleration due to gravity plus the constant acceleration. And so, for the higher acceleration I'm going to need a larger force. For larger mass I'm going to need a larger force. All makes physical sense. Okay. So here's our result. We have our angle theta, and the force required to have translation to the right. We did, we found that result by summing moments about an arbitrary point P, which was the point of contact on our body. Let's just double check to make sure that the sum of the moments About the mass center=0. So we're going to do a check here. So I've got some of the moments about C on the free body diagram= some of the moments about C on the kinetic diagram. I'll choose counter clockwise for assembling my equation. As my sine convention. And so the F force, will tend to cause a counter clockwise rotation about point C. So that's going to be positive in accordance with my sign convention. Its moment arm here is going to be L over 2, sine of theta. Now I have the N force. It's going to tend to cause a clockwise rotation about C, so that's going to be negative in accordance for my sign convention. So I got minus N X it's moment arm, and that moment arm is going to be. Here's the line of action, here's the point. Distance between the two is L over 2 cosine theta. And then I've mu N-force about point C. It's also going to cause a, going to cause a clock, a clockwise rotation which is negative in accordance to my [UNKNOWN] convention. So I have mu sub N times it's moment arm which is again, L/2 sin theta and that equals Since the effective force or the motion vector on my kinetic diagram, the line of action goes through point C. There's no moment there, so it's equal to 0. The moment arm is 0. And so I can now collect terms. Let's collect this first and third term here, so I get F. In fact before i even do that, let's cancel out the L, L over 2. That will make it a little easier over here. So I've got F X, or F - UN X sign theta, and that takes care of this term and this term. Equals, I'll move the, this term to the other side of the equation and I get N times cosine theta. But F minus mu N, if I put F minus mu N, that's equal to ma and N we said was equal to mg. And so I've got m, see this should be MA not MG. So I've got MAsinTheta = MGcosTheta. The Ms cancel and I get Tangent theta if I divide cosine is equal to G over A or A excuse me, theta is the inverse tangent of G over A and so, the sum of the moments about 0 does indeed check. And so, That's a good translation problem. I'd like you to try one on your own. Here's a worksheet for you to do. And I've got the solution in the handouts. And we'll see you next time.